You tie one end of 0.3-m-long spring to a 0.5 kg mass. Holding the other end of the spring in your hand, you whirl the mass in a vertical circle at a speed of 3 m/s.

Question

If the spring elongates 5 cm at the lowest point in the circular motion, what is the spring constant?

Answer 1

the centripetal force is
... (v^2/r) + (m g)

k * .05 = [3^2 / (.30 + .05)] + (.5 * 9.8)

the units are newtons per meter ... N/m